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# PUNJAB PUBLIC SERVICE COMMISSION WRITTEN TEST FOR THE POST OF LECTURER IN MATHEMATICS 2015

Differential Equation Related Material

PUNJAB PUBLIC SERVICE COMMISSION WRITTEN TEST FOR THE POST OF LECTURER IN MATHEMATICS 2015

Q.01 $\fn_cm&space;\int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=$__________________

A.   0

B.   Divergent

C.   -4

D.   4

C
View Explanation
$\fn_cm&space;\int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=-\frac{1}{2}\int_{-4}^{0}\frac{-2t}{\sqrt{16-t^2}}dt=-\sqrt{16-t^2}|_{-4}^{0}=-4$

Q.02  The period of function $\fn_cm&space;Acos\omega&space;t+Bsin\omega&space;t$ is

A.  $\fn_cm&space;\frac{\omega}{2\pi}$

B.  $\fn_cm&space;2\pi&space;\omega$

C.  $\fn_cm&space;\frac{2\omega}{\pi}$

D.  $\fn_cm&space;\frac{2\pi}{\omega}$

D
View Explanation

As sint and cost are periodic function with period 2π
So
Sinwt=sin(wt+2π)=sin(t+2π/w). Similarly for cost .
Hence period of given function if 2π/w
===> D is correct

Q.03 $\inline&space;\fn_cm&space;A=&space;(-4x-3y+az)i+(bx+3y+5z)j+(4x+cy+3z)k$ is irrotational if a,b,c are__________

A.   4,-3,5

B.   4,5,-3

C.   -3,4,5

D.   2,3,5

D
View Explanation

Flow is irrational if $\fn_cm&space;\nabla&space;\times&space;A=0\\&space;\\&space;\begin{bmatrix}&space;i&&space;j&space;&&space;k\\&space;\frac{\partial&space;}{\partial&space;x}&space;&&space;\frac{\partial&space;}{\partial&space;y}&space;&\frac{\partial&space;}{\partial&space;z}&space;\\&space;(-4x-3y+az)&&space;(bx+3y+5z)&space;&&space;(4x+cy+3z)&space;\end{bmatrix}=0\\&space;\\(c-5)i-(4-a)j+(b+3)k=0i+0j+0k\\&space;\\&space;(a,b,c)=(4,-3,5)$

Q.04 $\fn_cm&space;V=&space;(-4x-6y+3z)i+(-2x+y-5z)j+(5x+6y+az)k$  is solenoidal if a = ________

A.   1

B.   2

C.   3

4.   4

C
View Explanation

For solenoidal vector  $\fn_cm&space;\nabla$.V=0
After applying partial derivative we get
-4+1+a=0
So a=3
C is correct

Q.05 $\fn_cm&space;\fn_cm&space;\int_{(0,0)}^{(2,1)}&space;(10x^4-2xy^3)dx-3x^2y^2$    along the path $\fn_cm&space;x^4-6xy^3=4y^2&space;is$

A.   56

B.   60

C.   62

D.  64

B
View Explanation

We know that

$\fn_cm&space;M=(10x^4-2xy^3),N=-3x^2y^2\\&space;\\&space;\\&space;\frac{\partial&space;M}{\partial&space;y}=-6x^2\\\\\\&space;\frac{\partial&space;N}{\partial&space;x}=-6x^2$

As the solution is exact. So we have

$\fn_cm&space;\int_{(0,0)}^{(2,1)}&space;(10x^4-2xy^3)dx-independent&space;\;&space;terms&space;\;&space;of&space;\;&space;x\\\\\\&space;\int_{(0,0)}^{(2,1)}&space;(10x^4-2xy^3)dx-0&space;\\\\\\&space;={2x^5}-x^2y^3|_{(0,0)}^{(2,1)}\\\\\\&space;={64}-{4}=60$

Q.06 If S is the closed surface and v is the volume enclosed by S then $\fn_cm&space;\int&space;\int_s&space;r.nds=$
A. v
B.  2v
C.  3v
D.  4v

C
View Explanation

By Gauss Divergence Theorem

$\fn_cm&space;\int&space;\int_s&space;F.\widehat{N}ds=\int&space;\int_r&space;\int&space;dN.Fdv$

As position vector $\fn_cm&space;\overrightarrow{r}=&space;x&space;i+yj+zk$

So  $\fn_cm&space;\int&space;\int_s&space;r.nds=\int&space;\int_v&space;\int&space;\nabla.\overrightarrow{r}dv\\\\\\&space;=\int&space;\int_v&space;\int&space;(\frac{\partial&space;}{\partial&space;x}i+\frac{\partial&space;}{\partial&space;y}j+\frac{\partial&space;}{\partial&space;z}k).(xi+yj+zk)dv\\\\\\&space;=\int&space;\int_v&space;\int(1+1+1)dv\\\\\\&space;=3\int&space;\int_v&space;\int&space;dv&space;=3v$

Q.07 Centrifugal acceleration is

A.  $\fn_cm&space;-\omega&space;\times&space;(\omega&space;\times&space;r)$

B.  $\fn_cm&space;\omega&space;\times&space;(\omega&space;\times&space;r)$
C.  $\fn_cm&space;\omega&space;.&space;(\omega&space;\times&space;r)$
D.  $\fn_cm&space;r&space;\times&space;(\omega&space;\times&space;r)$

A
View Explanation

As $\fn_cm&space;\overrightarrow{r}=&space;cos&space;\omega&space;t&space;+&space;sin\omega&space;t\\\\\\&space;\frac&space;{d&space;\overrightarrow{r}}{dt}=\overrightarrow{v}=-&space;\omega&space;sin&space;\omega&space;t&space;+\omega&space;cos\omega&space;t\\\\\\&space;\frac&space;{d&space;\overrightarrow{v}}{dt}=\overrightarrow{a}=-&space;\omega^2&space;cos&space;\omega&space;t&space;-\omega^2&space;sin\omega&space;t\\\\\\&space;\overrightarrow{a}=-\omega^2&space;\overrightarrow{r}=-\omega&space;\times(&space;\omega&space;\times&space;\overrightarrow{r}&space;)$

Q.08  Number of degrees of freedom of two particles connected by a rigid rod moving treely in a plane Is

A. 2
B.  3
C.  4
D.  5

C
View Explanation

No. of degree of freedom= No. of particles × 2

= 2  ×  2  =   4

Q.09  The centrold of a uniform semicircular wire of radius a is

A. $\fn_cm&space;\frac{2a}{\pi}$

B.  $\fn_cm&space;\frac{4a}{\pi}$

C.  $\fn_cm&space;\frac{a}{\pi}$

D.  $\fn_cm&space;\frac{a}{2\pi}$

A
View Explanation

In processd

Q.10  Moment of inertia of a rectangular plate with sides a,b about an axis perpendicular to plate and passing through vertix is

A. $\fn_cm&space;\frac{1}{3}Ma^2$

B.  $\fn_cm&space;\frac{1}{3}Mb^2$

C. $\fn_cm&space;\frac{1}{3}M(a^2-b^2)$

D.  $\fn_cm&space;\frac{1}{3}M(a^2+b^2)$